Question

Calculate the pH of a 0.15 M lithium acetate (ka = 1.8 x 10 -5) .

Answer 1

Calculate the pH of a 0.15 M lithium acetate (ka = 1.8 x 10 -5) .

Initially, there is dissociatiuon of the salt, into cation / anions

LiAc --> Li+ + Ac-

Note that Ac- is a weak base, since it comes form a weak acid ( acetic acid )

then;

Ac- + H2O <-> HA + OH-

where:

Ac = acetate ion, HA = acetic acid, OH- hdyroxide ion, basic, and H2O = water

the pH can be calcualted as follows:

Kb = [HA][OH-]/[A-]

Note that Kb describes de basic equilibrium...

Kb can be calculated from Ka:

Kw = Ka*Kb

Kb = Kw/Ka = (10^-14)/(1.8*10^-5) = 5.5555*10^-10

Analysis of dissociaiton and concentrations

initially:

[HA] = 0

[OH-] = 0

[A-] = 0.15 M

in equilbirium

[HA] = 0 + x

[OH-] = 0 + x

[A-] = 0.15 - x

now, substitute in Kb

Kb = [HA][OH-]/[A-]

5.5555*10^-10 = (x*x) /(0.15 - x)

solve for x

x^2 + (5.55*10^-10)x - 0.15*(5.55*10^-10) = 0

x = 0.0000091238

now,

we know that

[OH-] = 0 + x = 0.0000091238

so

pOH = -log(OH) = -log(0.0000091238 = 5.0398

then

ph = 14-pOH = 14-5.0398

pH = 8.9602